Bridger Holden

2022-09-06

properties of logarithms $$\mathrm{ln}12-\mathrm{ln}2=\mathrm{ln}6$$

I checked wolframalpha and it says that ln12-ln2=ln6. How? i tried to do: ln12=ln(2*2*3) which may be 2ln(2*3) (which is probably wrong). I need help.

EDIT: Ok, thanks. Actually i could have just searched logarithms properties on google(didn't think about it). Sorry for taking your time.

I checked wolframalpha and it says that ln12-ln2=ln6. How? i tried to do: ln12=ln(2*2*3) which may be 2ln(2*3) (which is probably wrong). I need help.

EDIT: Ok, thanks. Actually i could have just searched logarithms properties on google(didn't think about it). Sorry for taking your time.

Krha77

Beginner2022-09-07Added 8 answers

You need to remember that

$$\mathrm{ln}a-\mathrm{ln}b=\mathrm{ln}\left(\frac{a}{b}\right)$$

Applying that here gives you

$$\mathrm{ln}(12)-\mathrm{ln}(2)=\mathrm{ln}\left(\frac{12}{2}\right)=\mathrm{ln}(6)$$

Note, alternatively, that we can use the property

$$\mathrm{ln}(ab)=\mathrm{ln}a+\mathrm{ln}b$$

as well.

$$\mathrm{ln}(12)=\mathrm{ln}(2\cdot 6)=\mathrm{ln}(2)+\mathrm{ln}6$$

So

$$\mathrm{ln}(12)-\mathrm{ln}2=\mathrm{ln}2+\mathrm{ln}6-\mathrm{ln}2=\mathrm{ln}6$$

$$\mathrm{ln}a-\mathrm{ln}b=\mathrm{ln}\left(\frac{a}{b}\right)$$

Applying that here gives you

$$\mathrm{ln}(12)-\mathrm{ln}(2)=\mathrm{ln}\left(\frac{12}{2}\right)=\mathrm{ln}(6)$$

Note, alternatively, that we can use the property

$$\mathrm{ln}(ab)=\mathrm{ln}a+\mathrm{ln}b$$

as well.

$$\mathrm{ln}(12)=\mathrm{ln}(2\cdot 6)=\mathrm{ln}(2)+\mathrm{ln}6$$

So

$$\mathrm{ln}(12)-\mathrm{ln}2=\mathrm{ln}2+\mathrm{ln}6-\mathrm{ln}2=\mathrm{ln}6$$

Sonia Rowland

Beginner2022-09-08Added 1 answers

Hint: Use the property

$$\mathrm{ln}x-\mathrm{ln}y=\mathrm{ln}\frac{x}{y}.$$

$$\mathrm{ln}x-\mathrm{ln}y=\mathrm{ln}\frac{x}{y}.$$

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